Question

1/You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:

HoHo : pA=0.25pA=0.25;  pB=0.4pB=0.4;  pC=0.1pC=0.1;  pD=0.25pD=0.25

Complete the table. Report all answers accurate to three decimal places.

Category	Observed Frequency	Expected Frequency
A	20
B	33
C	4
D	26

What is the chi-square test-statistic for this data? (2 decimal places)
χ2=( )
What is the P-Value? (3 decimal places)
P-Value = ( )
For significance level alpha 0.025,
What would be the conclusion of this hypothesis test?

• Reject the Null Hypothesis
• Fail to reject the Null Hypothesis
• Report all answers accurate to three decimal places

2/ You are conducting a multinomial hypothesis test (αα = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table.

Category	Observed Frequency	Expected Frequency
A	24
B	7
C	14
D	8
E	19

Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations.

What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.)
χ2=χ2=

What are the degrees of freedom for this test?
d.f.=

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null
• accept the alternative

As such, the final conclusion is that...

• A/ There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
• B/ There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
• C/ The sample data support the claim that all 5 categories are equally likely to be selected.
• D/ There is not sufficient sample evidence to support the claim that all 5 categories are equally likely to be selected.

Answer 1

1)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
A	0.2500	20.00	20.75	-0.16	0.027
B	0.4000	33.00	33.20	-0.03	0.001
C	0.1000	4.00	8.30	-1.49	2.228
D	0.2500	26.00	20.75	1.15	1.328
total	1.000	83	83		3.5843
test statistic X2 =		3.58
p value =		0.3100

Fail to reject the Null Hypothesis

2)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
A	0.2000	24.00	14.40	2.53	6.400
B	0.2000	7.00	14.40	-1.95	3.803
C	0.2000	14.00	14.40	-0.11	0.011
D	0.2000	8.00	14.40	-1.69	2.844
E	0.2000	19.00	14.40	1.21	1.469
total	1.000	72	72		14.5278
test statistic X2 =		14.528

degree of freedom =categories-1=

4

p value =

0.0058

  p-value is less than alpha

reject the null

• A/ There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.